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Question:

Let f: A → B be a function defined as f(x) = x² + x + 1, where A = ℝ - 2 and B = ℝ - 1. Then f is:

not invertible

invertible and f⁻¹(y) = 2y - y²

invertible and f⁻¹(y) = 2y + 1/y

invertible and f⁻¹(y) = 3y - y²

Solution:

Let y = f(x) ⇒ y = x² + x + 1 ⇒ yx - y = x² ⇒ x² - yx + y = 0. Solving for x using the quadratic formula, we have:
x = (y ± √(y² - 4y))/2
Since we have two possible values for x, this is not a one-to-one function, therefore, it is not invertible.